CHEM 4680 / Topic 3: pi

• When do you consider a ligand to be a pi-donor?

• When do you consider a ligand to be a pi-acceptor?

• If the p orbital on the uncoordinated ligand is occupied with two electrons, then the ligand is considered a pi-donor.

• If the orbitals on the uncoordinated ligands is vacant, the ligand is a pi-acceptor.

Give me five common pi donors.

• Amides.

• Alkoxides.

• Fluorides.

• Chlorides.

• Thiolates.

Explain how pi-donation works.

• Electron-deficient metals favour pi-donors because the metal ion orbitals (that are available for bonding overlap with pi-symmetric p orbitals) tend to be vacant.

• pi-donors donate electron density into the metal’s t_2g orbitals. These pi donors destabilize the t_2g orbitals. By raising this energy, it produces an MO bonding scheme wherein the t_2g* is now below the eg*, which gives rise to the new HOMO and LUMO.

• This decreases ∆_O. pi-donors are then classified as weak field ligands and hard bases.

What happens in the event of an octahedral d⁶ when put into a position of pi-donating?

The electrons brought cannot go into t_2g, so they will go to t_2g* instead. The complex will then be destabilized by the pi-donors due to the population of the antibonding t_2g*.

• What metals would inorganic amides and inorganic alkoxides often donate into?

• What metals would inorganic amides and inorganic alkoxides NOT often donate into?

• Briefly explain the difference between siloxides/aryloxides and alkoxides in regard to pi-donation.

• Early, electron-poor, high oxidation state metals are not as a good match because they have empty d-orbitals available for inorganic amides and inorganic alkoxides to donate into.

• Late, electron-rich metals, low oxidation state metals are not as good a match because of d-p repulsion. NOTE: The d orbitals are filled. If you gave electron-full d orbitals and p orbitals together → repulsion.

• Siloxides (R₃SiO^-) and aryloxides (ArO^-) are somewhat less donating ligands than plain alkoxides, as they are more stable than alkoxides on their own without the want to donate.

How can you spot the presence of pi-donation in alkylamides?

For alkylamides, pi-interactions are often easy to spot. Pi-interactions often lead to:

• Shorter bond distance between M-N.

• Restricted rotation about M-N.

• Less basicity of N.

Explain how pi-accepting works.

• Electron-rich metals favour pi-acceptors because the metal ion orbitals (that are available for bonding overlap with pi-symmetric p orbitals) tend to be full.

• pi-acceptors accept electron density from the metal’s t_2g orbitals. This stabilizes these said t_2g orbitals. By lowering this energy, it produces an MO bonding scheme wherein the t_2g* orbitals are now above eg*, and the t_2g is the new t_2g-symmetric set below the eg*, which gives rise to the new HOMO and LUMO.

• This increases ∆_O. pi-acceptors are then classified as strong field ligands and soft bases.

What happens in the event that there are no suitable electrons for pi-accepting?

Metals with no suitable electrons for back-donation (the result of pi-accepting) tend to form much weaker and less stable complexes with pi-acceptors.

What does the M-CO bond strength depend greatly on?

The M-CO bond strength depends on the amount of back-donation into the CO pi*.

Explain the three criteria for great back-donation into the pi antibonding orbitals of CO.

• If the metal centre is electron-rich and/or in a low oxidation state, (which means to have d electrons or have less d electrons removed), then CO will bind strongly.

• If the metal is not that electronegative (e.g. If the metal is very late, i.e. groups 11 and 12 or d⁹ and d¹⁰), because increased electronegativity and effective nuclear charge prohibits the metal from back-donating because it will not want to, even if the metal is electron-rich.

• If CO is not in competition with other pi-acceptors on the metal, the stronger the CO will bind because it has the metal’s undivided attention to receive its electrons from.

• Is there a possibility that the bonding for any CO ligand can be represented by resonance structures? Explain why or why not.

• What does increase in back donation into CO mean for oxygen in regards to charge?

• The bonding for any CO ligand may be represented by a series of resonance structures. We cannot reliably say (about any CO complex) that it corresponds to precisely to one of these resonance structures. What we can say is that, depending on how the metal binds (whether or not it forms one backbond or two backbonds), there can be a continuum of structures.

• Increasing back donation lowers the CO bond order and also increases the formal and actual charge on O and thus its basicity.

What IR absorptions correspond to in a molecule?

IR absorptions correspond to the energies of vibrations of the molecule that result in a net change of dipole moment.

• How does the bond stretching frequency relate to bond strength?

• What is the IR absorption of the triple bond in free CO?

• Would metal carbonyl complexes have lower or higher energy compared to bridging COs?

• Can you determine anything about a metal via the CO ligands via backbonding?

• For bond stretching frequencies, the energy of the stretching vibration is proportional to the bond strength via the force constant, i.e. the higher the energy of the bond stretching vibration, the stronger the bond.

• The triple bond in free CO has a stretch at 2143cm-1.

• Most metal carbonyl complexes absorb at lower energy because of back donation. Bridging COs typically have even lower IR frequencies (1720–1850 cm-1).

• CO ligands can then serve as a probe of how pi-basic and how electron-rich a metal centre is.

What does a wavenumber of 2100cm^-1 indicate for a metal in CO back bonding? Explain.

2100cm^-1 indicates a rather electron-poor metal centre, meaning it does not have electrons to give to the pi antibonding orbital. If it doesn’t have electrons to give, then the bond between C and O will remain a triple bond, which is high in energy.

What does a wavenumber of 1850cm^-1 indicate for a metal in CO back bonding? Explain.

Less than or equal to 1850cm^-1 indicates an electron-rich, pi-basic metal centre. If it does have electrons to give, then the bond between C and O can be broken due to the population of the antibonding orbital, reducing the bond order and making the bond lower in energy.

Explain the relationship between the electronegativity of a metal and wavenumber in CO back-bonding.

The more electronegative the metal centre, the higher the wavenumber because it wouldn’t want to give electrons to the antibonding orbital, which will NOT result in reducing the bond order, allowing the triple bond to stay.

Explain the relationship between the oxidation state on a metal and wavenumber in CO back-bonding.

• The more positive the charge or the more oxidized the metal is, the higher the wavenumber because if the metal is so oxidized, then that means there will be less electrons to get into the antibonding orbitals to reduce the bond order, letting the high-energy triple bond stay.

• The more negative the charge or the less oxidized the metal is, the lower the wavenumber because if the metal is not that oxidized and/or is rather reduced, then there will be more electrons to get into the antibonding orbitals to reduce the bond order, breaking that triple bond and reducing its energy.

Explain the relationship between the number of pi acids on a metal and wavenumber in CO back-bonding.

The less pi-acidic ligands bound to the metal, the lower the wavenumber. This is because if there are less pi-acidic ligands, then the CO will get more electron density and more back-donation. If there are more pi-acidic ligands, then the CO will get less electron density because the metal doesn’t give all of its electron density to just the one CO, it gives it to the other pi-acidic ligands.

Explain the why CN and R-NC is a stronger sigma donor than CO.

CN (cyanides) and R-NC (isonitriles) are stronger sigma-donors than CO. Compared to CO, nitrogen is less electronegative than oxygen. Because nitrogen is less electronegative, there is less polarization within CN. Because there is less polarization within CN, then there will be more electron density with the carbon atom. If there is more electron density for the carbon atom, which is where the sigma-bonding occurs, the stronger it is for sigma-donating.

Explain the why CN and R-NC is a weaker pi acceptor than CO.

CNs and R-NCs are weaker pi-acceptors than CO. Compared to CO, nitrogen is less electronegative than oxygen. The pi* formed between C and N is influenced because of the less electronegative nitrogen, rendering the pi* less electronegative. Because it is now less electronegative, then it will not want electrons that much, leading CN to be a weaker pi acceptor.

• What is the relationship between orbital overlap and stabilization?

• What is the relationship between energy difference between orbitals and stabilization?

• How can you explain the extra stabilization of the xy orbital in a square planar complex with CN as the ligands involved in pi-back donation?

• How can you explain the extra stabilization of the xz/yz orbital in a square planar complex with CN as the ligands involved in pi-back donation?

• As overlap increases, stabilization increases.

• As energy difference between parent orbitals increases, then stabilization decreases.

• The energy difference between the parent xy orbitals and the parent pi* orbitals is less than the energy difference between the parent yz/xz orbitals and the parent pi* orbitals.

  • Because the ∆E is less for xy-pi*, then xy will be stabilized more.

  • Because ∆E is more pronounced for yz/xz-pi, then yz/xz will be stabilized less.

• What will occur if isonitriles have electron-withdrawing R groups in regards to pi-accepting/pi-donating?

• What happens to isonitriles if the nitrogen is strongly reduced in regard to geometry? In what situations do you see nitrogens in isonitriles strongly reduced?

• If isonitriles have electron-withdrawing R groups, then the isonitrile will be more pi-accepting.

• If the nitrogen atom in isonitriles are strongly reduced, the R group will bend away and the bond angle will decrease. This bending effect is seen in complexes that have metals that are more electron-rich/electron-releasing (to donate into pi*) and ligands that have electron-withdrawing groups (to want more electrons into pi*).

Explain why NO is a stronger pi-acceptor than CO.

• NOs have a positive charge, which attracts the back-bonding d orbitals.

• NOs also have nitrogen, a more electronegative element than carbon, instead of carbon. Because nitrogen is more electronegative, this will influence the pi* in such a way that pi* will also be electronegative.

• What is the electron flow in linear and bent NO complexes? What form of the nitrosyl is present in each, and why?

• What explains the difference between the geometries in both complexes?

• The electron flow in linear NO complexes is NO to metal through a sigma bond. The nitrosyl is in its [NO]⁺ form because once you dissociate the sigma bond from the metal, you have a nitrosyl that is positively charged.

• The electron flow in bent NO complexes is metal to NO. NO is now a redox-active ligand, which has its nitrogen reduced two electrons by oxidizing the metal through an internal redox reaction. You can then consider [NO]⁺ to convert into [NO]^- because it is now reduced. Once it dissociates from the metal, it has now taken its electrons to become [NO]^-.

• The difference in geometry lies in the placement of lone pair on the bent structure. This pair causes a bent structure instead of linear because you now have a stereochemically active lone pair on the nitrogen.

If the geometry of a nitrosyl complex is dependent on the redox reaction, what determines if the nitrosyl internal redox reaction will occur, and how?

It will occur only if there is a reason for the loss of electrons, e.g. going from 20 electrons to 18 electrons (linear to bent because bent always takes electrons from the metal) or staying at 18 electrons and not going down to 16 electrons (preserving the linear geometry).

If you had a complex that had a linear NO but a relatively bond length for the linear NO bond and a relatively low wavenumber, what can you deduce?

A reduced NO, due to backbonding.

• Do electron counts involve sigma-ineractions?

• Do electron counts involve non-bonding ineractions?

• Do electron counts involve pi-ineractions?

Electron counts involves sigma-bonding interactions and the non-bonding electrons that do not participate in reactions, but it does not involve the pi-interactions, even if the pi-interactions involve significant electron density transfer to and from the metals.

What are the two interactions that an alkene can form? What results in a lengthening and weakening of the C-C bond in an alkene?

• Axial overlap, with cylindrical symmetry → sigma interactions. This uses the two electrons from alkene’s pi bond and takes them into the z² of the metal.

• Lateral overlap with restricted pi symmetry → pi interactions. This uses electron density from the metal centre to the pi*. This results in the lengthening and weakening of the C-C bond, i.e. double to single bond.

• What is an alkene complex?

• What is a metallocyclopropane (MCP) complex?

• What does the length of the C-C bond in alkene/MCP complexes say about the bond order and pi-back donation?

• As pi-basicity of M increases, what will form: alkene or MCP? Why?

• As EWG of R in CR₂=CR₂ increases, what will form: alkene or MCP? Why?

• A complex composed of a metal and alkene wherein the sigma-donation dominates.

• A complex composed of a metal and alkene wherein pi-back donation dominates.

• The longer the C-C bond, the more single bond-like, the more pi-back donation there is.

• As pi-basicity of M increases, the more MCP it is. This is because it has electron density to give to the ethylene for pi-back donation.

• As EWG of R on CR₂=CR₂ increases, the more MCP it is. This is because it would want more electron density from the metal for pi-back donation.

How does the electron-accepting/releasing nature of a ligand on a complex with an alkene influence the bond length of the alkene and whether or not the complex will be in the alkene extreme or MCP extreme? Explain why.

Replacing an electron-accepting ligand with an electron-releasing ligand, this will lead to an increase in pi-back donation and thus lengthening of the C-C bond. This is because the electron-releasing ligand will donate electron density to the metal, which the metal can give as a back donation to the ethylene.

How does electronegativity and oxidation state of the metal centre affect bond length? Explain why.

• The less electronegative the metal, the stronger the back bonding, lengthening the C-C bond. This is because if the metal is not that electron-hungry (less EN).

• If it is more reduced than oxidized (negative or less positive OS), then it will want to give electrons to the metal more.

NOTE: Comparison is best with identical d systems but with different electronegativities and oxidation state.

Among the rows of transition metals, which row of transition metals leads to MCP? What is the reason for this?

There is stronger back donation with third row metals, leading to a lengthening of C-C bond. This is because of the better overlap that is made by larger and more diffuse d orbitals.

What does the “bending away” of the substituents mean for an alkene that can either go into alkene- or MCP-type complex? Explain.

• If the complex has the substituents of CR₂=CR₂ strongly bent away, then it is most likely to be in the MCP range.

• Carbons become nearly sp³ hybridized due to the back bonding.

• As the bond angles at the carbon change, the substituents will be more bent away.

• What happens in regard to reactivity if the sigma donation dominates? Why does this happen?

• What happens in regard to reactivity if the pi-back donation dominates? Why does this happen?

• If the sigma donation dominates, the alkene is depleted of electron density (due to giving its electrons to the metal) and is susceptible to nucleophilic attack.

• If the pi-back donation dominates, the two ends of the double bond become carbanionic in character (due to the electrons it receives from the metal) and is susceptible to electrophilic attack.